1.已知向量a=(sinQ,1),b=(1,cosQ),-π/22.已知ABC为三角形ABC三内角,向量m=(-1,根号3),n=(cosA,sinA),且m×n=1若(1+sin2B)/(cos^2B-sin^2B)=-3,求谈B,tanC(cos^2B是cosB的平方)
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![1.已知向量a=(sinQ,1),b=(1,cosQ),-π/22.已知ABC为三角形ABC三内角,向量m=(-1,根号3),n=(cosA,sinA),且m×n=1若(1+sin2B)/(cos^2B-sin^2B)=-3,求谈B,tanC(cos^2B是cosB的平方)](/uploads/image/z/5168820-12-0.jpg?t=1.%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28sinQ%2C1%29%2Cb%3D%281%2CcosQ%29%2C-%CF%80%2F22.%E5%B7%B2%E7%9F%A5ABC%E4%B8%BA%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%89%E5%86%85%E8%A7%92%EF%BC%8C%E5%90%91%E9%87%8Fm%3D%EF%BC%88-1%EF%BC%8C%E6%A0%B9%E5%8F%B73%EF%BC%89%EF%BC%8Cn%3D%EF%BC%88cosA%EF%BC%8CsinA%EF%BC%89%EF%BC%8C%E4%B8%94m%C3%97n%3D1%E8%8B%A5%281%2Bsin2B%29%2F%28cos%5E2B-sin%5E2B%29%3D-3%2C%E6%B1%82%E8%B0%88B%2CtanC%28cos%5E2B%E6%98%AFcosB%E7%9A%84%E5%B9%B3%E6%96%B9%29)
1.已知向量a=(sinQ,1),b=(1,cosQ),-π/22.已知ABC为三角形ABC三内角,向量m=(-1,根号3),n=(cosA,sinA),且m×n=1若(1+sin2B)/(cos^2B-sin^2B)=-3,求谈B,tanC(cos^2B是cosB的平方)
1.已知向量a=(sinQ,1),b=(1,cosQ),-π/2
2.已知ABC为三角形ABC三内角,向量m=(-1,根号3),n=(cosA,sinA),且m×n=1
若(1+sin2B)/(cos^2B-sin^2B)=-3,求谈B,tanC
(cos^2B是cosB的平方)
1.已知向量a=(sinQ,1),b=(1,cosQ),-π/22.已知ABC为三角形ABC三内角,向量m=(-1,根号3),n=(cosA,sinA),且m×n=1若(1+sin2B)/(cos^2B-sin^2B)=-3,求谈B,tanC(cos^2B是cosB的平方)
(1)a+b=(sinQ+1,1+cosQ)
|a+b|=√(a+b)²=√((sinQ+1)²+(1+cosQ)²)=√(sin²Q+cos²Q+2(sinQ+cosQ)+2)
=√(2(sinQ+cosQ)+3)
=√(2√2sin(Q+π/4)+3)
- π/4 <Q+π/4<3π/4 ∴-√2/2<2sin(Q+π/4)≤1 ∴1<|a+b|≤√(2√2+3)=√(1+2√2+2)=1+√2
(2)
m×n=1
-cosA+√3sinA=1 √3sinA=1+cosA两边平方,3sin^2A=(1+cosA)² sin^2A=1-2cosA+cos^2B
解得cosA=1/2,∴A=60°
(1+sin2B)/(cos^2B-sin^2B)=-3
(1+sin2B)/(cos2B)=-3
同理可求cos2B=0,B=45°tanB=1 C=75°
tan75°=tan(45°+30°)=(tan45+tan30)/(1-tan45*tan30)=2+√3.